Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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Geometry - triangle in a circle
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Brent@GMATPrepNow
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Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
Brent Hanneson - Creator of GMATPrepNow.com
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Brent@GMATPrepNow
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Righto.
(1) INSUFF – No idea about the length of BC
(2) INSUFF – No idea about the other sides
(1) and (2) – If the radius=6 and BC=12, then BC is the diameter. If B is the diameter, then angle BAC is 90 degrees.
We can let x=AC=AB and create the equation x^2 + x^2 = 12^2
We could solve this for x and, subsequently, find the area of the triangle.
SUFF
Answer: C
(1) INSUFF – No idea about the length of BC
(2) INSUFF – No idea about the other sides
(1) and (2) – If the radius=6 and BC=12, then BC is the diameter. If B is the diameter, then angle BAC is 90 degrees.
We can let x=AC=AB and create the equation x^2 + x^2 = 12^2
We could solve this for x and, subsequently, find the area of the triangle.
SUFF
Answer: C
Brent Hanneson - Creator of GMATPrepNow.com
Bern,Brent Hanneson wrote:Righto.
(1) INSUFF – No idea about the length of BC
(2) INSUFF – No idea about the other sides
(1) and (2) – If the radius=6 and BC=12, then BC is the diameter. If B is the diameter, then angle BAC is 90 degrees.
We can let x=AC=AB and create the equation x^2 + x^2 = 12^2
We could solve this for x and, subsequently, find the area of the triangle.
SUFF
Answer: C
i have a question here.
is it possible to draw a right traingle such that diameter is not one of the sides of the triangle??
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Brent@GMATPrepNow
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If you're referring to triangles in circles in general, the anwer is no.is it possible to draw a right traingle such that diameter is not one of the sides of the triangle??
See:
Brent Hanneson - Creator of GMATPrepNow.com
Then, why is A insufficient for this problem...Brent Hanneson wrote:If you're referring to triangles in circles in general, the anwer is no.is it possible to draw a right traingle such that diameter is not one of the sides of the triangle??
See:
if the angle subtended by the chord is 90, then that chord is the diameter.
since we have radius in the question we have all required info to solve this problem...
am i missing something?
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Brent@GMATPrepNow
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Statement (1) provides no indication that there is a 90 degree angle in the triangle.aroon7 wrote:Then, why is A insufficient for this problem...Brent Hanneson wrote:If you're referring to triangles in circles in general, the anwer is no.is it possible to draw a right traingle such that diameter is not one of the sides of the triangle??
See:
if the angle subtended by the chord is 90, then that chord is the diameter.
since we have radius in the question we have all required info to solve this problem...
am i missing something?
Brent Hanneson - Creator of GMATPrepNow.com
i was fooled by the diagram...Brent Hanneson wrote:Statement (1) provides no indication that there is a 90 degree angle in the triangle.aroon7 wrote:Then, why is A insufficient for this problem...Brent Hanneson wrote:If you're referring to triangles in circles in general, the anwer is no.is it possible to draw a right traingle such that diameter is not one of the sides of the triangle??
See:
if the angle subtended by the chord is 90, then that chord is the diameter.
since we have radius in the question we have all required info to solve this problem...
am i missing something?
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vikram_k51
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1:AC=AB
Not Sufficient.We don't know whether,BC is the diameter.It can be a simple chord as well.
2.BC=12;
Implies ABC is a right angled trianlge.
Area of a right triangle=1/2*a*b(AC=a;AB=b;)
We know:a^2+b^2=144.But this is not sufficient.
Together with A,the soln is Sufficient.
Hence C
Not Sufficient.We don't know whether,BC is the diameter.It can be a simple chord as well.
2.BC=12;
Implies ABC is a right angled trianlge.
Area of a right triangle=1/2*a*b(AC=a;AB=b;)
We know:a^2+b^2=144.But this is not sufficient.
Together with A,the soln is Sufficient.
Hence C
